TITLE: Prove that $2^n3^{2n}-1$ is always divisible by 17
QUESTION [7 upvotes]: Prove that $2^n3^{2n} -1$ is always divisible by $17$.
I am very new to proofs and i was considering using proof by induction but I am not sure how to. I know you have to start by verifying the statement is true for the integer 1 but I dont know where to go from there.

REPLY [3 votes]: Hint: $2\cdot3^2=18$ and $18\equiv1\pmod{17}$
If you do not know that
$$
a\equiv b\pmod{p}\implies a^n=b^n\pmod{p}\tag{1}
$$
then $(1)$ can be proven by induction pretty easily.